Since \(u\in \mathcal {V}^{1,p}(D)\) , we have \(u^{\langle p/2 \rangle }\in W^{1,2}(D)\) . By Maz'ya [1]} we get that \(\nabla u^{\langle p/2 \rangle }\)  exists in the classical sense almost everywhere and coincides with the weak gradient. Formula (REF ) in case \(p\ge 2\)  now follows from the chain rule, thus in the sequel we assume that \(p\in (1,2)\) . Note that if \(\nabla u^{\langle p/2 \rangle }(x)\) 
exists in the classical sense, then either \(u(x)\ne 0\)  or \(u(x) = 0\)  and \(\nabla u(x) = 0\) . Indeed, let \(e_i\)  be the unit vector in the \(i\) -th coordinate direction and assume that \(u(x)=0\)  and \(\partial _i u(x) \ne 0\) . Then
\(\bigg |\frac{u^{\langle p/2 \rangle }(x+he_i) - u^{\langle p/2 \rangle }(x)}{h}\bigg | = \bigg |\frac{u^{\langle p/2 \rangle }(x+he_i)}{h}\bigg |\gtrsim |h|^{p/2-1} \mathop {\longrightarrow }\limits _{h\rightarrow 0} \infty ,\) 
which contradicts the existence of \(\nabla u^{\langle p/2 \rangle }(x)\) .
Thus by Lemma REF  we get that (REF ) holds for a.e. \(x\in D\) .
